# Project Management - Part 2

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Project Management - Part 2

Overview What happens when activity times in a project are: Not fixed
Crashing the project Not known with certainty PERT (Project Evaluation and Review Technique)

Estimated time is fixed... no more.
Project Crashing Estimated time is fixed... no more. Can reduce the length of time of a project through additional resources manpower, equipment Direct cost of activity is always increased

Project Crashing: Motivation
Need to reduce time of project because: Requirement to complete in specified time frame Economic advantage Three kinds of costs Crash costs (activity direct costs) Administration costs (or project indirect costs) Penalty costs Incur crash costs to avoid administration and penalty costs

Calculating Crash Costs per Unit Time
9000 – 3000 Crash Cost / Unit Time = 7 – 3 Crash Cost = \$9,000 = \$1,500 Normal Cost = \$3,000 Crash Time =3 Normal Time =7

Project Crashing: Procedure
Always crash one period at a time! Identify critical activities Select least expensive to crash Savings? Implement if so. Update all paths Repeat 1-3 until no cost savings are possible

Which activity should we crash?
5 3 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350 Y-\$125 Z-\$325 Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 2 2 1 Admin Cost = \$500 Path Duration ABCD AXYZ 16 15 Which activity should we crash?

A B C D X Y Z 3 3 2 4 5 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350
Normal Duration: (per period) Minimum Duration: 2 1 Admin Cost = \$500 Path Duration ABCD 16 AXYZ 15

A B C D X Y Z 5 3 3 2 4 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350
Normal Duration: (per period) Minimum Duration: 2 1 Admin Cost = \$500 Path Duration ABCD 16 AXYZ 15

A B C D X Y Z 3 5 3 2 4 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350
Normal Duration: (per period) Minimum Duration: 2 1 Admin Cost = \$500 Path Duration ABCD 16 AXYZ 15

3 5 5 3 A B C D X Y Z 3 2 4 5 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350 Y-\$125 Z-\$325 Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 1 2 2 Admin Cost = \$500 2 1 B Path Duration ABCD 16 AXYZ 15 Choose B because it is the cheapest of the three alternatives (A, B, or C)

Which activity(s) should we crash next?
3 4 3 A B C D X Y Z 3 2 4 5 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350 Y-\$125 Z-\$325 Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 2 2 1 Admin Cost = \$500 2 1 B Path Duration ABCD AXYZ 100 500 400 Cost Save Net Cumul Which activity(s) should we crash next?

A B C D X Y Z B C D A X Y Z B 4 4 3 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 2 1 Admin Cost = \$500 2 2 1 B Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B 4 3 4 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 2 1 2 2 1 Admin Cost = \$500 B Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B 4 3 4 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 1 2 2 2 1 Admin Cost = \$500 B Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B 4 3 4 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 2 2 1 Admin Cost = \$500 2 1 B Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B 3 4 4 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 1 2 2 Admin Cost = \$500 2 1 B Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B 3 4 4 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z Admin Cost = \$500 1 2 2 2 1 B Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B 3 4 4 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 1 2 2 2 1 Admin Cost = \$500 B Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B BY 3 4 4 3 3 2 4 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 4 3 5 Minimum Duration: 2 X Y Z 1 2 2 2 1 Admin Cost = \$500 B BY Path Duration ABCD AXYZ Cost Save Net Cumul

Which activity(s) should we crash next?
3 3 3 A B C D X Y Z 3 2 5 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350 Y-\$125 Z-\$325 Normal Duration: B C D (per period) 3 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 2 2 1 Admin Cost = \$500 2 1 B BY Path Duration ABCD AXYZ Cost Save Net Cumul Which activity(s) should we crash next?

A B C D X Y Z B C D A X Y Z B BY 3 3 3 3 3 2 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 2 2 Admin Cost = \$500 2 1 1 B BY Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B BY 3 3 3 3 3 2 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 2 2 1 Admin Cost = \$500 2 1 B BY Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B BY 3 3 3 2 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 2 Admin Cost = \$500 2 1 B BY Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B BY 3 3 3 3 3 2 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 1 2 2 Admin Cost = \$500 2 1 B BY Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B BY A 3 3 3 3 3 2 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 3 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 1 2 2 Admin Cost = \$500 2 1 B BY A Path Duration ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B BY A 3 3 3 2 5 5 Crash Costs A -\$250
Normal Duration: B C D (per period) 2 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 2 Admin Cost = \$500 2 1 B BY A Path Duration ABCD AXYZ Cost Save Net Cumul

Which activity(s) should we crash next?
3 3 A B C D X Y Z 3 2 5 5 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350 Y-\$125 Z-\$325 Normal Duration: B C D (per period) 2 2 3 A 3 3 3 5 Minimum Duration: 2 X Y Z 2 Admin Cost = \$500 2 1 B BY A Path Duration ABCD AXYZ Cost Save Net Cumul Which activity(s) should we crash next?

A B C D X Y Z B C D A X Y Z B BY A CY 3 3 3 2 5 4 4 Crash Costs
Normal Duration: B C D (per period) 2 2 3 A 3 2 3 5 Minimum Duration: 2 X Y Z 2 Admin Cost = \$500 2 1 B BY A CY Path Duration ABCD AXYZ Cost Save Net Cumul

Which activity(s) should we crash next?
3 3 A B C D X Y Z 3 2 5 4 4 Crash Costs A -\$250 B-\$100 C-\$200 D- X-\$350 Y-\$125 Z-\$325 Normal Duration: B C D (per period) 2 2 3 A 3 2 3 5 Minimum Duration: 2 X Y Z 2 Admin Cost = \$500 2 1 B BY A CY Path Duration ABCD AXYZ Cost Save Net Cumul Which activity(s) should we crash next?

We lose \$\$, so do not crash
3 3 A B C D X Y Z 3 2 5 3 Normal Duration: B C D 2 2 3 A 3 2 2 5 Minimum Duration: 2 X Y Z 2 Admin Cost = \$500 2 1 B BY A CY CZ Path Duration We lose \$\$, so do not crash CZ ABCD AXYZ Cost Save Net Cumul

A B C D X Y Z B C D A X Y Z B BY A CY 3 3 3 2 5 4 4 2 2 3 3 2 3 5 2 2
Normal Duration: B C D 2 2 3 A 3 2 3 5 Minimum Duration: 2 X Y Z 2 2 1 Achieved by crashing B BY A CY Path Duration Most economical duration ABCD AXYZ Cost Save Net Cumul Spent \$900 in increased direct costs Avoided \$2000 in administration costs Net savings

PERT (CPM With 3 Time Estimates)
In past, time estimate is firm Now… task duration is uncertain New activity Natural variance Use 3 time estimates to deal with uncertainty a = most optimistic estimate m = most likely b = most pessimistic

et (expected time) = (a + 4m + b)/6 σ (standard dev) = (b - a)/6
Formulas et (expected time) = (a + 4m + b)/6 σ (standard dev) = (b - a)/6 σ 2 (variance) = (b - a)2/36 = [(b - a)/6]2 Z = (D - et)/σ How many standard deviations away D is from et

Activity List for Example Problem
Immediate Required Activity Activity Description Predecessors Time (weeks) A Select office site - 3 B Create organization and financial plan - 5 C Determine personnel requirements B 3 D Design facility A,C 4 E Construct the interior D 8 F Select personnel to move C 2 G Hire new employees F 4 H Move records, key personnel, etc. F 2 I Make financial arrangements B 5 J Train new personnel H,E,G 3

Mean and Variance et = (a + 4m + b)/6 σ = (b - a)/6
For activity A, et = (1+4·3+5)/6 s = (5-1)/6 Mean and Variance s2 = [(5-1)/6]2 Activity a m b et ET ó ó 2 A 1 3 5 3 2/3 4/9 B 3 4.5 9 5 1 1 C 2 3 4 3 1/3 1/9 D 2 4 6 4 2/3 4/9 E 4 7 16 8 2 4 F 1 1.5 5 2 2/3 4/9 G 2.5 3.5 7.5 4 5/6 25/36 H 1 2 3 2 1/3 1/9 I 4 5 6 5 1/3 1/9 J 1.5 3 4.5 3 1/2 1/4 et = (a + 4m + b)/6 σ = (b - a)/6 σ 2 = (b - a)2/36 = [(b - a)/6]2

Path Length Path length is the sum of expected times of activities on the path not sum of most likely times Longest expected path length is critical Path length is uncertain, and so is project duration Path has variance equal to sum of variances of individual activities on path

A D E G J B C F H I

A D E G J B C F H I

Calculating probability of different completion time
If we want to know the probability of completing the project in 22 weeks or less: 0.4 0.3 0.2 0.1 22 et = 23

Take .42 to the Z table => 1-.66 = .34
Calculations et = 23 weeks σ2CP =s2BCDEJ Z = (D - et)/σ CP = (22-23)/ = -0.42 Take .42 to the Z table => = .34 Beware - never add standard deviations! = σ 2B + σ 2C + σ 2D + σ 2E + σ 2J =

16% chance between 22 and 23 34% chance < 22 (table)
22 is 0.42 SD below Mean of 23 16% chance between 22 and 23 0.4 0.3 0.2 0.1 34% chance < 22 (table) 50% chance > 23 22 et = 23

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